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The Complete Equilibrium

A rod of mass...

Question

A rod of mass $m$ is placed on two supports as shown. Calculate the normal reaction of each support.

N_{1} = \frac{1}{3} m g, N_{2} = \frac{2}{3} m g

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N_{1} = \frac{2}{3} m g, N_{2} = \frac{1}{3} m g

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N_{1} = \frac{m g}{2}, N_{2} = \frac{m g}{2}

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N_{1} = \frac{m g}{4}, N_{2} = \frac{3 m g}{4}

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Solution

The correct option is A $N_{1} = \frac{1}{3} m g, N_{2} = \frac{2}{3} m g$
As th rod is in equilibrium,
$∵ τ_{A} = 0$
$∴ (N_{1} \times 0) + (N_{2} \times \frac{3 L}{4}) - (m g \times \frac{L}{2}) = 0$
$N_{2} = \frac{2}{3} m g$
As, $N_{1} + N_{2} = m g$
and $N_{1} = \frac{1}{3} m g$

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